Holt McDougal Algebra 2 8-2 Multiplying and Dividing Rational Expressions 8-2 Multiplying and...

Holt McDougal Algebra 2

8-2 Multiplying and Dividing Rational Expressions8-2 Multiplying and Dividing Rational Expressions

Holt Algebra 2

Warm Up

Lesson Presentation

Lesson Quiz



8-2 Multiplying and Dividing Rational Expressions

Warm UpSimplify each expression. Assume all variables are nonzero.1. x5 x2

3.

x7

x6

x2

Factor each expression.

5. x2 – 2x – 8

6. x2 – 5x

(x – 4)(x + 2)

x3(x – 3)(x + 3)

2. y3 y3 y6

x4 y2

y54. 1

y3

7. x5 – 9x3

x(x – 5)



Simplify rational expressions.

Multiply and divide rational expressions.

Objectives



rational expression

Vocabulary



In Lesson 8-1, you worked with inverse variation functions such as y = . The expression on the right side of this equation is a rational expression. A rational expression is a quotient of two polynomials. Other examples of rational expressions include the following:

5x



When identifying values for which a rational expression is undefined, identify the values of the variable that make the original denominator equal to 0.

Caution!

Because rational expressions are ratios of polynomials, you can simplify them the same way as you simplify fractions. Recall that to write a fraction in simplest form, you can divide out common factors in the numerator and denominator.



Simplify. Identify any x-values for which the expression is undefined.

Example 1A: Simplifying Rational Expressions

Quotient of Powers Property

10x8

6x4

510x8 – 4

3653

x4=

The expression is undefined at x = 0 because this value of x makes 6x4 equal 0.



Example 1B: Simplifying Rational Expressions


x2 + x – 2 x2 + 2x – 3

(x + 2)(x – 1) (x – 1)(x + 3)

Factor; then divide out common factors.

= (x + 2)(x + 3)

The expression is undefined at x = 1 and x = –3 because these values of x make the factors (x – 1) and (x + 3) equal 0.



Check Substitute x = 1 and x = –3 into the original expression.

(1)2 + (1) – 2 (1)2 + 2(1) – 3

00

= (–3)2 + (–3) – 2 (–3)2 + 2(–3) – 3

40=

Both values of x result in division by 0, which is undefined.

Example 1B Continued



Check It Out! Example 1a


Quotient of Powers Property

16x11

8x2

28x11 – 2

18 2x9=

The expression is undefined at x = 0 because this value of x makes 8x2 equal 0.



Check It Out! Example 1b


3x + 4 3x2 + x – 4

(3x + 4) (3x + 4)(x – 1)


= 1(x – 1)

The expression is undefined at x = 1 and x = –because these values of x make the factors (x – 1) and (3x + 4) equal 0.

43



3(1) + 43(1)2 + (1) – 4

70

=


Check It Out! Example 1b Continued

Check Substitute x = 1 and x = – into the original expression.

43



Check It Out! Example 1c


6x2 + 7x + 2 6x2 – 5x – 5

(2x + 1)(3x + 2) (3x + 2)(2x – 3)


= (2x + 1)(2x – 3)

The expression is undefined at x =– and x =

because these values of x make the factors (3x + 2)

and (2x – 3) equal 0.

32

23




Check Substitute x = and x = – into the original expression.

32

23

Check It Out! Example 1c Continued



Simplify . Identify any x values

for which the expression is undefined.

Example 2: Simplifying by Factoring by –1

Factor out –1 in the numerator so that x2 is positive, and reorder the terms.

Factor the numerator and denominator. Divide out common factors.

The expression is undefined at x = –2 and x = 4.

4x – x2 x2 – 2x – 8

–1(x2 – 4x)x2 – 2x – 8

–1(x)(x – 4)(x – 4)(x + 2)

–x(x + 2 ) Simplify.



Check The calculator screens suggest that

= except when x = – 2

or x = 4.

Example 2 Continued

4x – x2 x2 – 2x – 8

–x(x + 2)






Factor out –1 in the numerator so that x is positive, and reorder the terms.


The expression is undefined at x = 5.

10 – 2x x – 5

–1(2x – 10)x – 5

–2 1 Simplify.

–1(2)(x – 5)(x – 5)



Check It Out! Example 2a Continued


= –2 except when x = 5.10 – 2x x – 5






Factor out –1 in the numerator so that x is positive, and reorder the terms.


–x2 + 3x 2x2 – 7x + 3

–1(x2 – 3x)2x2 – 7x + 3

–x 2x – 1 Simplify.

–1(x)(x – 3)(x – 3)(2x – 1)

The expression is undefined at x = 3 and x = . 1 2




–x2 + 3x 2x2 – 7x + 3


= except when x =

and x = 3.

–x 2x – 1

12



You can multiply rational expressions the same way that you multiply fractions.



Multiply. Assume that all expressions are defined.

Example 3: Multiplying Rational Expressions

A. 3x5y3 2x3y7

10x3y4 9x2y5

3x5y3 2x3y7

10x3y4 9x2y5

53

3

5x3 3y5

B. x – 3 4x + 20

x + 5x2 – 9

x – 3 4(x + 5)

x + 5(x – 3)(x + 3)

1 4(x + 3)



Check It Out! Example 3

Multiply. Assume that all expressions are defined.

A. x15

20x4

2x x7

x15

20x4

2x x7

3

22

2x3 3

B. 10x – 40 x2 – 6x + 8

x + 35x + 15

10(x – 4) (x – 4)(x – 2)

x + 35(x + 3)

2 (x – 2)

2



You can also divide rational expressions. Recall that to divide by a fraction, you multiply by its reciprocal.

1 2

3 4

÷ = 1 2

4 3

2 2

3=



Divide. Assume that all expressions are defined.

Example 4A: Dividing Rational Expressions

Rewrite as multiplication by the reciprocal.

5x4

8x2y2÷

8y5

15

5x4

8x2y2

158y5

5x4

8x2y2

158y5

3

2 3

x2y3 3



Example 4B: Dividing Rational Expressions

x4 – 9x2 x2 – 4x + 3

÷ x4 + 2x3 – 8x2

x2 – 16


x4 – 9x2 x2 – 4x + 3

x2 – 16

x4 + 2x3 – 8x2


x2 (x2 – 9)x2 – 4x + 3

x2 – 16

x2(x2 + 2x – 8)

x2(x – 3)(x + 3)(x – 3)(x – 1)

(x + 4)(x – 4)x2(x – 2)(x + 4)

(x + 3)(x – 4) (x – 1)(x – 2)



x4 y



x2

4÷

12y2

x4y

x2

4 12y2

2

3y x2


x2

4

x4y12y2

3 1




2x2 – 7x – 4 x2 – 9

÷ 4x2– 18x2 – 28x +12


(2x + 1)(x – 4)(x + 3)(x – 3)

4(2x2 – 7x + 3)(2x + 1)(2x – 1)

(2x + 1)(x – 4)(x + 3)(x – 3)

4(2x – 1)(x – 3)(2x + 1)(2x – 1)

4(x – 4) (x +3)

2x2 – 7x – 4 x2 – 9

8x2 – 28x +124x2– 1



Example 5A: Solving Simple Rational Equations

Solve. Check your solution.

Note that x ≠ 5.

x2 – 25x – 5

= 14

(x + 5)(x – 5)(x – 5)

= 14

x + 5 = 14

x = 9



x2 – 25x – 5

= 14Check

(9)2 – 259 – 5

14

56 4

14

14 14

Example 5A Continued



Example 5B: Solving Simple Rational Equations


Note that x ≠ 2.

x2 – 3x – 10 x – 2

= 7

(x + 5)(x – 2)(x – 2)

= 7

x + 5 = 7

x = 2Because the left side of the original equation is undefined when x = 2, there is no solution.



Check A graphing calculator shows that 2 is not a solution.

Example 5B Continued





Note that x ≠ –4.

x2 + x – 12 x + 4

= –7

(x – 3)(x + 4)(x + 4)

= –7

x – 3 = –7

x = –4Because the left side of the original equation is undefined when x = –4, there is no solution.



Check It Out! Example 5a Continued

Check A graphing calculator shows that –4 is not a solution.




4x2 – 92x + 3

= 5

(2x + 3)(2x – 3)(2x + 3)

= 5

2x – 3 = 5

x = 4


Note that x ≠ – .32



4x2 – 92x + 3

= 5Check

4(4)2 – 92(4) + 3

5

55 11

5

5 5




Lesson Quiz: Part I

1.

2.


x2 – 6x + 5 x2 – 3x – 10

6x – x2

x2 – 7x + 6

x – 1 x + 2

x ≠ –2, 5

–x x – 1

x ≠ 1, 6



Lesson Quiz: Part II

3.

Multiply or divide. Assume that all expressions are defined.

x2 + 4x + 3x2 – 4

÷ x2 + 2x – 3x2 – 6x + 8

4.

x + 1 3x + 6

6x + 12x2 – 1

(x + 1)(x – 4) (x + 2)(x – 1)

2 x – 1

5. x = 4


4x2 – 12x – 1

= 9

Holt McDougal Algebra 2 8-2 Multiplying and Dividing Rational Expressions 8-2 Multiplying and...

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